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Math Algebra

posted by on .

A quick question to check my answer.

Trains A & B are travelling in the same direction on parallel tracks. Train A is travelling at 100 miles per hour and train B is travelling at 110 miles per hour. Train A passes a station at 2:10 A.M. If train B passes the same station at 2:40 A.M., at what time will train B catch up to train A?

My answer is 7:40 A.M.

Here is my reasoning:

Train A is ahead of Train B by 30 minutes.
This means that by the time train B gets to the station, train A would have travelled 50 miles.

So every hour that Train A travels, Train B closes the gap because it is 10 miles an hour faster.

If I start their time at 2:40 A.M., here is what I come up with:

Time/Train-A/Train-B
2:40/50/0
3:40/150/110
4:40/250/220
5:40/350/330
6:40/450/440
7:40/550/550 <-- Both trains would have travelled the same distance by 7:40.

Can someone double check. Thanks.

  • Math Algebra - ,

    your answer is correct

    an easier way is the following:

    let the time past 2:10 be t hours

    so when they pass each other , the slower train has gone for t hours and the faster train for 1-1/2 hrs.

    so 110(t-1/2) = 100t
    10 t = 55
    t = 5.5 hrs or 5 hours and 30 min

    add that to 2:10 to get 7:40

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